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Re: Problem with change of variables in an integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg131574] Re: Problem with change of variables in an integral
  • From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>
  • Date: Thu, 5 Sep 2013 08:13:38 -0400 (EDT)
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Hello,

Although I know how to make a change of variables in an integral I can only do
it manually by applying a substitution rule to the integrand and the
differential e.g
                       {f[z],\[DifferentialD]z}//. {z-> r E^(I
\[Phi]),\[DifferentialD]z->E^(I \[Phi]) \[DifferentialD]r,\[Phi] -> (2\[Pi]/3}

But it cannot applied this substitution rule directly to the integral, e.g.
                         Integrate[f[z],{z,0,\[Infinity]}] //. {z-> r E^(I
\[Phi]),\[DifferentialD]z->E^(I \[Phi]) \[DifferentialD]r,\[Phi] -> (2\[Pi]/3}

Comparing with the correct result, the exponential factor E^((2 I \[Pi])/3) =
(-1)^(2/3) is missing in the evaluation of the integral. The correct appearance
of the
integral is :  Integrate[1/(1+r^3) E^((2 I \[Pi])/3),{r,0,\[Infinity]}]

How can I force Mathematica (V8) to perform the correct transformation of
variables as regards to the integral (and not to its separate parts of it as
{f[z],\[DifferentialD]z} ?

Any suggestions are appreciated.
Robert Kragler

Hello, Robert,

Try this:

Clear[f, \[Phi]];
Map[ReplaceAll[#, {z -> r*Exp[I \[Phi]] , \[Phi] -> 2 \[Pi]/3}] &,
 Integrate[f[z], {z, 0, \[Infinity]}]]

to map the replacements directly onto the structures below the Integrate operator. The result is below:

\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Infinity]\)]\(f[
\*SuperscriptBox[\(E\), \(I\ \[Phi]\)]\ r] \[DifferentialD]\((
\*SuperscriptBox[\(E\), \(I\ \[Phi]\)]\ r)\)\)\)

If it is used, for example, with the function f[z_]:=1/(1+z^3), as you tried:

Clear[f, \[Phi]];
f[z_] := 1/(1 + z^3);
Map[ReplaceAll[#, {z -> r*Exp[I \[Phi]] , \[Phi] -> 2 \[Pi]/3}] &,
 Integrate[f[z], {z, 0, \[Infinity]}]]

it gives

(2 \[Pi])/(3 Sqrt[3])

When I do the same "by hand":

\[Phi] = 2 \[Pi]/3;
Integrate[Exp[I \[Phi]]/(
 1 + (Exp[I \[Phi]]*r)^3), {r, 0, \[Infinity]}]

I get this:

(2 (-1)^(2/3) \[Pi])/(3 Sqrt[3])

which in my opinion is almost the same, though not entirely. I do not see, where from the difference comes.

Surprisingly strange result has been delivered, when I used Esc+e+e+Esc instead of Exp to check of how the "hand-made" transformation would work:

\[Phi] = 2 \[Pi]/3;
Integrate[E^(i \[Phi])/(1 + (E^(i \[Phi]) r)^3), {r, 0, \[Infinity]}]

The answer was:

ConditionalExpression[(2 E^((2 i \[Pi])/3) \[Pi])/(
 3 Sqrt[3] (E^(2 i \[Pi]))^(1/3)),
 Im[E^(-2 i \[Pi])] != 0 || Re[E^(-2 i \[Pi])] >= 0]

which I do not understand. I do it on PC, Win XP, Mathematica 9.0.1.0. I would be v=
ery interested in comments of the Community on this matter.

Best, Alexei



Alexei BOULBITCH, Dr., habil.
IEE S.A.
ZAE Weiergewan,
11, rue Edmond Reuter,
L-5326 Contern, LUXEMBOURG

Office phone :  +352-2454-2566
Office fax:       +352-2454-3566
mobile phone:  +49 151 52 40 66 44

e-mail: alexei.boulbitch at iee.lu






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